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Question Number 215141 by cherokeesay last updated on 29/Dec/24

	Answered by aleks041103 last updated on 30/Dec/24
	$eqn: (x−s)^2 +y^2 =r^2 the following systems must have one soln { ((y=(√x))),(((x−s)^2 +y^2 =r^2 )) :} and { ((4x+4y=17)),(((x−s)^2 +y^2 =r^2 )) :} also 0<s<17/4 1st: x^2 +(1−2s)x+s^2 −r^2 =0 ⇒one soln if (1−2s)^2 −4(s^2 −r^2 )=0 ⇒1+4s^2 −4s−4s^2 +4r^2 =0 ⇒4r^2 −4s+1=0 ⇒s−(1/4)=r^2 ⇒eqn: (x−s)^2 +y^2 =s−(1/4) 4x+4y=17 is tangent to the circle ⇒the tangent is st 45° ⇒s+(√2)r=((17)/4) ⇒(1/4)+r^2 +(√2)r=((17)/4)⇒r^2 +(√2)r−4=0 r_(1/2) =((−(√2)±(√(2+16)))/2)=((√2)/2)(−1±3) r>0⇒r=(√2) ⇒S=(1/2)πr^2 =π ⇒S=π, (x−(9/4))^2 +y^2 =2 Alternayively: (x−(1/4)−(s−1/4))^2 +y^2 =(s−1/4) ⇒4(x−1/4)+4y=16 ⇒(x−(1/4))+y=4 z=x−1/4 ⇒(z−r^2 )^2 +y^2 =r^2 and z+y=4 ⇒(z−r^2 )^2 +(4−z)^2 =r^2 ⇒z^2 −2zr^2 −r^2 +r^4 +16−8z+z^2 =0 ⇒2z^2 −2(r^2 +4)z+16−r^2 +r^4 =0 one soln for z iff (r^2 +4)^2 +2(−16−r^4 +r^2 )=0 ⇒r^4 +8r^2 +16−32−2r^4 +2r^2 =0 ⇒−r^4 +10r^2 −16=0⇒r^4 −10r^2 +16=0 ⇒r^2 =((10±(√(10^2 −4.16)))/2)=((10±6)/2)=5±3=2,8 x<((17)/4)⇒z<4 ⇒ r^2 =2$
	$e q n : {(x - s)}^{2} + y^{2} = r^{2}$ $t h e f o l l o w i n g s y s t e m s m u s t h a v e o n e s o l n$ ${\begin{cases} y = \sqrt{x} \\ {(x - s)}^{2} + y^{2} = r^{2} \end{cases} a n d {\begin{cases} 4 x + 4 y = 17 \\ {(x - s)}^{2} + y^{2} = r^{2} \end{cases}$ $a l s o 0 < s < 17 / 4$ $1 s t : x^{2} + (1 - 2 s) x + s^{2} - r^{2} = 0$ $\Rightarrow o n e s o l n i f {(1 - 2 s)}^{2} - 4 (s^{2} - r^{2}) = 0$ $\Rightarrow 1 + 4 s^{2} - 4 s - 4 s^{2} + 4 r^{2} = 0$ $\Rightarrow 4 r^{2} - 4 s + 1 = 0 \Rightarrow s - \frac{1}{4} = r^{2}$ $\Rightarrow e q n : {(x - s)}^{2} + y^{2} = s - \frac{1}{4}$ $4 x + 4 y = 17 i s t a n g e n t t o t h e c i r c l e$ $\Rightarrow t h e t a n g e n t i s s t 45 °$ $\Rightarrow s + \sqrt{2} r = \frac{17}{4}$ $\Rightarrow \frac{1}{4} + r^{2} + \sqrt{2} r = \frac{17}{4} \Rightarrow r^{2} + \sqrt{2} r - 4 = 0$ $r_{1 / 2} = \frac{- \sqrt{2} \pm \sqrt{2 + 16}}{2} = \frac{\sqrt{2}}{2} (- 1 \pm 3)$ $r > 0 \Rightarrow r = \sqrt{2}$ $\Rightarrow S = \frac{1}{2} π r^{2} = π$ $\Rightarrow S = π, {(x - \frac{9}{4})}^{2} + y^{2} = 2$ $A l t e r n a y i v e l y :$ ${(x - \frac{1}{4} - (s - 1 / 4))}^{2} + y^{2} = (s - 1 / 4)$ $\Rightarrow 4 (x - 1 / 4) + 4 y = 16$ $\Rightarrow (x - \frac{1}{4}) + y = 4$ $z = x - 1 / 4$ $\Rightarrow {(z - r^{2})}^{2} + y^{2} = r^{2} a n d z + y = 4$ $\Rightarrow {(z - r^{2})}^{2} + {(4 - z)}^{2} = r^{2}$ $\Rightarrow z^{2} - 2 {z r}^{2} - r^{2} + r^{4} + 16 - 8 z + z^{2} = 0$ $\Rightarrow 2 z^{2} - 2 (r^{2} + 4) z + 16 - r^{2} + r^{4} = 0$ $o n e s o l n f o r z i f f$ ${(r^{2} + 4)}^{2} + 2 (- 16 - r^{4} + r^{2}) = 0$ $\Rightarrow r^{4} + 8 r^{2} + 16 - 32 - 2 r^{4} + 2 r^{2} = 0$ $\Rightarrow - r^{4} + 10 r^{2} - 16 = 0 \Rightarrow r^{4} - 10 r^{2} + 16 = 0$ $\Rightarrow r^{2} = \frac{10 \pm \sqrt{10^{2} - 4.16}}{2} = \frac{10 \pm 6}{2} = 5 \pm 3 = 2, 8$ $x < \frac{17}{4} \Rightarrow z < 4 \Rightarrow r^{2} = 2$
	Commented by aleks041103 last updated on 30/Dec/24

	Commented by cherokeesay last updated on 30/Dec/24

	$s o n i c e!$ $t h a n k y o u m a s t e r!$
	Answered by mr W last updated on 30/Dec/24

	Commented by mr W last updated on 30/Dec/24

	$i n t h e r o t a t e d c o o r d i n a t e s y s t e m :$ $y = x^{2}$ $s a y P (p, p^{2})$ $\tan θ = \frac{d y}{d x} = 2 p$ $x_{C} = p - r \sin θ = 0$ $\Rightarrow p = r \sin θ = \frac{2 p r}{\sqrt{1 + 4 p^{2}}}$ $\Rightarrow 1 = \frac{2 r}{\sqrt{1 + 4 p^{2}}} \Rightarrow p = \sqrt{r^{2} - \frac{1}{4}}$ $y_{C} = p^{2} + r \cos θ = p^{2} + \frac{r}{\sqrt{1 + 4 p^{2}}}$ $= r^{2} - \frac{1}{4} + \frac{1}{2} = r^{2} + \frac{1}{4}$ $x_{Q} = - \frac{r}{\sqrt{2}}$ $y_{Q} = r^{2} + \frac{1}{4} + \frac{r}{\sqrt{2}}$ $- 4 (- \frac{r}{\sqrt{2}}) + 4 (r^{2} + \frac{1}{4} + \frac{r}{\sqrt{2}}) = 17$ $r^{2} + \sqrt{2} r - 4 = 0$ $r = \frac{- \sqrt{2} + 3 \sqrt{2}}{2} = \sqrt{2} ✓$ $a r e a o f s e m i - c i r c l e = \frac{π r^{2}}{2} = π ✓$
	Commented by cherokeesay last updated on 30/Dec/24

	$g r e a t j o b!$ $t h a n k s o m u c h s i r!$
	Commented by mr W last updated on 31/Dec/24

	Commented by mr W last updated on 31/Dec/24

	$y = \sqrt{x} \Rightarrow x = y^{2}$ $P (p^{2}, p)$ $\tan θ = \frac{d x}{d y} = 2 p$ $p = r \sin θ = \frac{2 p r}{\sqrt{1 + 4 p^{2}}} \Rightarrow p^{2} = r^{2} - \frac{1}{4}$ $x_{C} = p^{2} + r \cos θ = p^{2} + \frac{r}{\sqrt{1 + 4 p^{2}}} = r^{2} - \frac{1}{4} + \frac{1}{2} = r^{2} + \frac{1}{4}$ $x_{C} = \frac{17}{4} - \sqrt{2} r$ $r^{2} + \frac{1}{4} = \frac{17}{4} - \sqrt{2} r$ $r^{2} + \sqrt{2} r - 4 = 0$ $\Rightarrow r = \sqrt{2}$
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